Two-Dimensional Motion of a Particle: Velocity & Acceleration

In summary, the two-dimensional motion of a particle is described by the equations r = 1/(sin\theta - cos\theta) and tan\theta = 1 + 1/t^2, where r and \theta are measured in meters and radians, respectively, and t is measured in seconds. To find the magnitudes of velocity and acceleration at any instant, we can differentiate r and \theta to get \dot{r}, \ddot{r}, \dot{\theta}, and \ddot{\theta}. However, it is not necessary to solve for \theta, as we can use a right triangle to find expressions for sin\theta and cos\theta in terms of t. This will also help us find
  • #1
chart2006
12
0

Homework Statement



The two-dimensional motion of a particle is defined by the relationship [tex] r = \frac {1}{sin\theta - cos\theta} [/tex] and [tex] tan\theta = 1 + \frac {1}{t^2} [/tex], where [tex] r [/tex] and [tex] \theta [/tex] are expressed in meters and radians, respectively, and [tex] t [/tex] is expressed in seconds. Determine (a) the magnitudes of velocity and acceleration at any instant, (b) the radius of curvature of the path.


Homework Equations



[tex] r = \frac {1}{sin\theta - cos\theta} [/tex]


[tex] tan\theta = 1 + \frac {1}{t^2} [/tex]



The Attempt at a Solution



I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex] tan\theta [/tex] for [tex] \theta [/tex]. Once I've done that I figure I'd need to differentiate both [tex] r [/tex] and [tex] \theta [/tex] to find [tex] \dot{r}, \ddot{r}, \dot{\theta}, \ddot{\theta}[/tex].

I don't know if I'm on the correct route but any help would be appreciated. thanks!
 
Physics news on Phys.org
  • #2
chart2006 said:
I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex] tan\theta [/tex] for [tex] \theta [/tex].

Tangent is not a one-to-one function, so that's a bad idea.

Instead, draw a picture!:smile: I think you can find expressions for [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex] in terms of [itex]t[/itex] without actually solving for [itex]\theta[/itex] first...think 'right triangle':wink:
 

1. What is the difference between velocity and acceleration?

Velocity is the rate of change of an object's displacement over time, while acceleration is the rate of change of an object's velocity over time. In other words, velocity tells us how fast and in what direction an object is moving, while acceleration tells us how much an object's velocity is changing.

2. How do you calculate the velocity of a particle?

To calculate the velocity of a particle, you need to divide the change in its position (displacement) by the change in time. This is represented by the formula: velocity = change in position / change in time.

3. How is acceleration represented in a position-time graph?

In a position-time graph, acceleration is represented by the slope of the line. A steeper slope indicates a higher acceleration, while a flatter slope indicates a lower acceleration or a constant velocity.

4. Can a particle have a constant velocity and non-zero acceleration?

Yes, a particle can have a constant velocity and non-zero acceleration. This can occur when the particle is moving in a circular path, as the direction of its velocity is constantly changing, leading to a non-zero acceleration.

5. What is the difference between average acceleration and instantaneous acceleration?

Average acceleration is calculated over a period of time, while instantaneous acceleration is calculated at a specific moment in time. Average acceleration gives us an overall idea of how an object's velocity is changing, while instantaneous acceleration gives us the exact rate of change at a particular instant.

Similar threads

Replies
6
Views
929
  • Advanced Physics Homework Help
Replies
4
Views
304
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
883
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
923
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
14
Views
2K
Back
Top